<br>

<i>This program was contributed by Jörg Frohne (University of Siegen,
Germany) while on a long-term visit to Texas A&amp;M University.
<br>
This material is based upon work partly supported by ThyssenKrupp Steel Europe.
</i>


<a name="Intro"></a>
<h3>Introduction</h3>

This example is based on the Laplace equation in 2d and deals with the
question what happens if a membrane is deflected by some external force but is
also constrained by an obstacle. In other words, think of a elastic membrane
clamped at the boundary to a rectangular frame (we choose $\Omega =
\left[-1,1\right]^2$) and that sags through due to gravity acting on it. What
happens now if there is an obstacle under the membrane that prevents it from
reaching its equilibrium position if gravity was the only existing force? In
the current example program, we will consider that under the membrane is a
stair step obstacle against which gravity pushes the membrane.

This problem is typically called the "obstacle problem" (see also <a
href="http://en.wikipedia.org/wiki/Obstacle_problem">this Wikipedia article</a>), and it results in a
variational inequality, rather than a variational equation when put into the
weak form. We will below derive it from the classical formulation, but before we
go on to discuss the mathematics let us show how the solution of the problem we
will consider in this tutorial program looks to gain some intuition of what
we should expect:

<table align="center" class="tutorial" cellspacing="3" cellpadding="3">
  <tr>
    <td align="center">
        <img src="https://www.dealii.org/images/steps/developer/step-41.displacement.png" alt="">
    </td>
    <td align="center">
        <img src="https://www.dealii.org/images/steps/developer/step-41.active-set.png" alt="">
    </td>
  </tr>
</table>

Here, at the left, we see the displacement of the membrane. The shape
of the obstacle underneath is clearly visible. On the right, we overlay which
parts of the membrane are in contact with the obstacle. We will later call
this set of points the "active set" to indicate that an inequality constraint
is active there.


<h3>Classical formulation</h3>

The classical formulation of the problem possesses the following form:
@f{align*}
 -\textrm{div}\ \sigma &\geq f & &\quad\text{in } \Omega,\\
 \sigma &= \nabla u & &\quad\text{in } \Omega,\\
 u(\mathbf x) &= 0 & &\quad\text{on }\partial\Omega,\\
(-\Delta u - f)(u - g) &= 0 & &\quad\text{in } \Omega,\\
 u(\mathbf x) &\geq g(\mathbf x) & &\quad\text{in } \Omega
@f}
with $u\in H^2(\Omega)$.  $u$ is a scalar valued function that denotes the
vertical displacement of the membrane. The first equation is called equilibrium
condition with a force of areal density $f$. Here, we will consider this force
to be gravity. The second one is known as Hooke's Law that says that the stresses
$\sigma$ are proportional to the gradient of the displacements $u$ (the
proportionality constant, often denoted by $E$, has been set to one here,
without loss of generality; if it is constant, it can be put into the right
hand side function). At the boundary we have zero Dirichlet
conditions. Obviously, the first two equations can be combined to yield
$-\Delta u \ge f$.

Intuitively, gravity acts downward and so $f(\mathbf x)$ is a negative
function (we choose $f=-10$ in this program). The first condition then means
that the total force acting on the membrane is gravity plus something
positive: namely the upward force that the obstacle exerts on the membrane at
those places where the two of them are in contact. How big is this additional
force? We don't know yet (and neither do we know "where" it actually acts) but
it must be so that the membrane doesn't penetrate the obstacle.

The fourth equality above together with the last inequality forms the obstacle
condition which has to hold at every point of the whole domain. The latter of
these two means that the membrane must be above the obstacle $g(\mathbf x)$
everywhere. The second to last equation, often called the "complementarity
condition" says that where the membrane is not in contact with the obstacle
(i.e., those $\mathbf x$ where $u(\mathbf x) - g(\mathbf x) \neq 0$), then
$-\Delta u=f$ at these locations; in other words, no additional forces act
there, as expected. On the other hand, where $u=g$ we can have $-\Delta u-f
\neq 0$, i.e., there can be additional forces (though there don't have to be:
it is possible for the membrane to just touch, not press against, the
obstacle).


<h3>Derivation of the variational inequality</h3>

An obvious way to obtain the variational formulation of the obstacle problem is to consider the total potential energy:
@f{equation*}
 E(u) \dealcoloneq \dfrac{1}{2}\int\limits_{\Omega} \nabla u \cdot \nabla u - \int\limits_{\Omega} fu.
@f}
We have to find a solution $u\in G$ of the following minimization problem:
@f{equation*}
 E(u)\leq E(v)\quad \forall v\in G,
@f}
with the convex set of admissible displacements:
@f{equation*}
 G \dealcoloneq \lbrace v\in V: v\geq g \text{ a.e. in } \Omega\rbrace,\quad V\dealcoloneq H^1_0(\Omega).
@f}
This set takes care of the third and fifth conditions above (the boundary
values and the complementarity condition).

Consider now the minimizer $u\in G$ of $E$ and any other function $v\in
G$. Then the function
@f{equation*}
 F(\varepsilon) \dealcoloneq E(u+\varepsilon(v-u)),\quad\varepsilon\in\left[0,1\right],
@f}
takes its minimum at $\varepsilon = 0$ (because $u$ is a minimizer of the
energy functional $E(\cdot)$), so that $F'(0)\geq 0$ for any choice
of $v$. Note that
$u+\varepsilon(v-u) = (1-\varepsilon)u+\varepsilon v\in G$ because of the
convexity of $G$. If we compute $F'(\varepsilon)\vert_{\varepsilon=0}$ it
yields the variational formulation we are searching for:

<i>Find a function $u\in G$ with</i>
@f{equation*}
 \left(\nabla u, \nabla(v-u)\right) \geq \left(f,v-u\right) \quad \forall v\in G.
@f}

This is the typical form of variational inequalities, where not just $v$
appears in the bilinear form but in fact $v-u$. The reason is this: if $u$ is
not constrained, then we can find test functions $v$ in $G$ so that $v-u$ can have
any sign. By choosing test functions $v_1,v_2$ so that $v_1-u = -(v_2-u)$ it
follows that the inequality can only hold for both $v_1$ and $v_2$ if the two
sides are in fact equal, i.e., we obtain a variational equality.

On the other hand, if $u=g$ then $G$ only allows test functions $v$ so that in fact
$v-u\ge 0$. This means that we can't test the equation with both $v-u$ and
$-(v-u)$ as above, and so we can no longer conclude that the two sides are in
fact equal. Thus, this mimics the way we have discussed the complementarity
condition above.



<h3>Formulation as a saddle point problem</h3>

The variational inequality above is awkward to work with. We would therefore
like to reformulate it as an equivalent saddle point problem. We introduce a
Lagrange multiplier $\lambda$ and the convex cone $K\subset V'$, $V'$
dual space of $V$, $K \dealcoloneq \{\mu\in V': \langle\mu,v\rangle\geq 0,\quad \forall
v\in V, v \le 0 \}$ of
Lagrange multipliers, where $\langle\cdot,\cdot\rangle$ denotes the duality
pairing between $V'$ and $V$. Intuitively, $K$ is the cone of all "non-positive
functions", except that $K\subset (H_0^1)'$ and so contains other objects
besides regular functions as well.
This yields:

<i>Find $u\in V$ and $\lambda\in K$ such that</i>
@f{align*}
 a(u,v) + b(v,\lambda) &= f(v),\quad &&v\in V\\
 b(u,\mu - \lambda) &\leq \langle g,\mu - \lambda\rangle,\quad&&\mu\in K,
@f}
<i>with</i>
@f{align*}
 a(u,v) &\dealcoloneq \left(\nabla u, \nabla v\right),\quad &&u,v\in V\\
 b(u,\mu) &\dealcoloneq \langle u,\mu\rangle,\quad &&u\in V,\quad\mu\in V'.
@f}
In other words, we can consider $\lambda$ as the negative of the additional, positive force that the
obstacle exerts on the membrane. The inequality in the second line of the
statement above only appears to have the wrong sign because we have
$\mu-\lambda<0$ at points where $\lambda=0$, given the definition of $K$.

The existence and uniqueness of $(u,\lambda)\in V\times K$ of this saddle
point problem has been stated in Glowinski, Lions and Tr&eacute;moli&egrave;res: Numerical Analysis of Variational
Inequalities, North-Holland, 1981.



<h3>Active Set methods to solve the saddle point problem</h3>

There are different methods to solve the variational inequality. As one
possibility you can understand the saddle point problem as a convex quadratic program (QP) with
inequality constraints.

To get there, let us assume that we discretize both $u$ and $\lambda$ with the
same finite element space, for example the usual $Q_k$ spaces. We would then
get the equations
@f{eqnarray*}
 &A U + B\Lambda = F,&\\
 &[BU-G]_i \geq 0, \quad \Lambda_i \leq 0,\quad \Lambda_i[BU-G]_i = 0
\qquad \forall i.&
@f}
where $B$ is the @ref GlossMassMatrix "mass matrix" on the chosen finite element space and the
indices $i$ above are for all degrees of freedom in the set $\cal S$ of degrees of
freedom located in the interior of the domain
(we have Dirichlet conditions on the perimeter). However, we
can make our life simpler if we use a particular quadrature rule when
assembling all terms that yield this mass matrix, namely a quadrature formula
where quadrature points are only located at the interpolation points at
which shape functions are defined; since all but one shape function are zero
at these locations, we get a diagonal mass matrix with
@f{align*}
  B_{ii} = \int_\Omega \varphi_i(\mathbf x)^2\ \textrm{d}x,
  \qquad
  B_{ij}=0 \ \text{for } i\neq j.
@f}
To define $G$ we use the same technique as for $B$. In other words, we
define
@f{align*}
  G_{i} = \int_\Omega g_h(x) \varphi_i(\mathbf x)\ \textrm{d}x,
@f}
where $g_h$ is a suitable approximation of $g$. The integral in the definition
of $B_{ii}$ and $G_i$ are then approximated by the trapezoidal rule.
With this, the equations above can be restated as
@f{eqnarray*}
 &A U + B\Lambda = F,&\\
 &U_i-B_{ii}^{-1}G_i \ge 0, \quad \Lambda_i \leq 0,\quad \Lambda_i[U_i-B_{ii}^{-1}G_i] = 0
\qquad \forall i\in{\cal S}.&
@f}

Now we define for each degree of freedom $i$ the function
@f{equation*}
 C([BU]_i,\Lambda_i) \dealcoloneq -\Lambda_i + \min\lbrace 0, \Lambda_i + c([BU]_i - G_i) \rbrace,
@f}
with some $c>0$. (In this program we choose $c = 100$. It is a kind of a
penalty parameter which depends on the problem itself and needs to be chosen
large enough; for example there is no convergence for $c = 1$ using the
current program if we use 7 global refinements.)

After some head-scratching one can then convince oneself that the inequalities
above can equivalently be rewritten as
@f{equation*}
 C([BU]_i,\Lambda_i) = 0, \qquad \forall i\in{\cal S}.
@f}
The primal-dual active set strategy we will use here is an iterative scheme which is based on
this condition to predict the next active and inactive sets $\mathcal{A}_k$ and
$\mathcal{F}_k$ (that is, those complementary sets of indices $i$ for which
$U_i$ is either equal to or not equal to the value of the obstacle
$B^{-1}G$). For a more in depth treatment of this approach, see Hintermueller, Ito, Kunisch: The primal-dual active set
strategy as a semismooth newton method, SIAM J. OPTIM., 2003, Vol. 13, No. 3,
pp. 865-888.

<h3>The primal-dual active set algorithm</h3>

The algorithm for the primal-dual active set method works as follows (NOTE: $B = B^T$):

1. Initialize $\mathcal{A}_k$ and $\mathcal{F}_k$, such that
 $\mathcal{S}=\mathcal{A}_k\cup\mathcal{F}_k$ and
 $\mathcal{A}_k\cap\mathcal{F}_k=\emptyset$ and set $k=1$.
2. Find the primal-dual pair $(U^k,\Lambda^k)$ that satisfies
 @f{align*}
  AU^k + B\Lambda^k &= F,\\
  [BU^k]_i &= G_i\quad&&\forall i\in\mathcal{A}_k,\\
  \Lambda_i^k &= 0\quad&&\forall i\in\mathcal{F}_k.
 @f}
 Note that the second and third conditions imply that exactly $|S|$ unknowns
 are fixed, with the first condition yielding the remaining $|S|$ equations
 necessary to determine both $U$ and $\Lambda$.
3. Define the new active and inactive sets by
 @f{equation*}
 \begin{split}
  \mathcal{A}_{k+1} \dealcoloneq \lbrace i\in\mathcal{S}:\Lambda^k_i + c([BU^k]_i - G_i)< 0\rbrace,\\
  \mathcal{F}_{k+1} \dealcoloneq \lbrace i\in\mathcal{S}:\Lambda^k_i + c([BU^k]_i - G_i)\geq 0\rbrace.
 \end{split}
 @f}
4. If $\mathcal{A}_{k+1}=\mathcal{A}_k$ (and then, obviously, also
 $\mathcal{F}_{k+1}=\mathcal{F}_k$) then stop, else set $k=k+1$ and go to step
 (2).

The method is called "primal-dual" because it uses both primal (the
displacement $U$) as well as dual variables (the Lagrange multiplier
$\Lambda$) to determine the next active set.

At the end of this section, let us add two observations. First,
for any primal-dual pair $(U^k,\Lambda^k)$ that satisfies these
condition, we can distinguish the following cases:

1. $\Lambda^k_i + c([BU^k]_i - G_i) < 0$ (i active):
  <br>
  Then either $[BU^k]_i<G_i$ and $\Lambda^k_i=0$ (penetration) or $\Lambda^k_i<0$ and $[BU^k]_i=G_i$ (pressing load).
2. $\Lambda^k_i + c([BU^k]_i - G_i)\geq 0$ (i inactive):
  <br>
  Then either $[BU^k]_i\geq G_i$ and $\Lambda^k_i=0$ (no contact) or $\Lambda^k_i\geq0$ and $[BU^k]_i=G_i$ (unpressing load).

Second, the method above appears intuitively correct and useful but a bit ad
hoc. However, it can be derived in a concisely in the following way. To this
end, note that we'd like to solve the nonlinear system
@f{eqnarray*}
 &A U + B\Lambda = F,&\\
 &C([BU-G]_i, \Lambda_i) = 0,
\qquad \forall i.&
@f}
We can iteratively solve this by always linearizing around the previous
iterate (i.e., applying a Newton method), but for this we need to linearize
the function $C(\cdot,\cdot)$ that is not differentiable. That said, it is
slantly differentiable, and in fact we have
@f{equation*}
 \dfrac{\partial}{\partial U^k_i}C([BU^k]_i,\Lambda^k_i) = \begin{cases}
                                   cB_{ii},& \text{if}\ \Lambda^k_i + c([BU^k]_i - G_i)< 0\\
                                   0,& \text{if}\ \Lambda^k_i + c([BU^k]_i - G_i)\geq 0.
                                  \end{cases}
@f}
@f{equation*}
 \dfrac{\partial}{\partial\Lambda^k_i}C([BU^k]_i,\Lambda^k_i) = \begin{cases}
                                   0,& \text{if}\ \Lambda^k_i + c([BU^k]_i - G_i)< 0\\
                                   -1,& \text{if}\ \Lambda^k_i + c([BU^k]_i - G_i)\geq 0.
                                  \end{cases}
@f}
This suggest a semismooth Newton step of the form
@f{equation*}
 \begin{pmatrix}
 A_{\mathcal{F}_k\mathcal{F}_k} & A_{\mathcal{F}_k\mathcal{A}_k} & B_{\mathcal{F}_k} & 0\\
 A_{\mathcal{A}_k\mathcal{F}_k} & A_{\mathcal{A}_k\mathcal{A}_k} & 0 & B_{\mathcal{A}_k}\\
 0 & 0 & -Id_{\mathcal{F}_k} & 0\\
 0 & cB_{\mathcal{A}_k} & 0 & 0
\end{pmatrix}
\begin{pmatrix}
 \delta U^k_{\mathcal{F}_k}\\ \delta U^k_{\mathcal{A}_k}\\ \delta \Lambda^k_{\mathcal{F}_k}\\ \delta \Lambda^k_{\mathcal{A}_k}
\end{pmatrix}
=
-\begin{pmatrix}
 (AU^k + \Lambda^k - F)_{\mathcal{F}_k}\\ (AU^k + \Lambda^k - F)_{\mathcal{A}_k}\\ -\Lambda^k_{\mathcal{F}_k}\\ c(B_{\mathcal{A}_k} U^k - G)_{\mathcal{A}_k}
\end{pmatrix},
@f}
where we have split matrices $A,B$ as well as vectors in the natural way into
rows and columns whose indices belong to either the active set
${\mathcal{A}_k}$ or the inactive set ${\mathcal{F}_k}$.

Rather than solving for updates $\delta U, \delta \Lambda$, we can also solve
for the variables we are interested in right away by setting $\delta U^k \dealcoloneq
U^{k+1} - U^k$ and $\delta \Lambda^k \dealcoloneq \Lambda^{k+1} - \Lambda^k$ and
bringing all known terms to the right hand side. This yields
@f{equation*}
\begin{pmatrix}
 A_{\mathcal{F}_k\mathcal{F}_k} & A_{\mathcal{F}_k\mathcal{A}_k} & B_{\mathcal{F}_k} & 0\\
 A_{\mathcal{A}_k\mathcal{F}_k} & A_{\mathcal{A}_k\mathcal{A}_k} & 0 & B_{\mathcal{A}_k}\\
 0 & 0 & Id_{\mathcal{F}_k} & 0\\
 0 & B_{\mathcal{A}_k} & 0 & 0
\end{pmatrix}
\begin{pmatrix}
 U^k_{\mathcal{F}_k}\\ U^k_{\mathcal{A}_k}\\ \Lambda^k_{\mathcal{F}_k}\\ \Lambda^k_{\mathcal{A}_k}
\end{pmatrix}
=
\begin{pmatrix}
 F_{\mathcal{F}_k}\\ F_{\mathcal{A}_k}\\ 0\\ G_{\mathcal{A}_k}
\end{pmatrix}.
@f}
These are the equations outlined above in the description of the basic algorithm.

We could even drive this a bit further.
It's easy to see that we can eliminate the third row and the third column
because it implies $\Lambda_{\mathcal{F}_k} = 0$:
@f{equation*}
\begin{pmatrix}
 A_{\mathcal{F}_k\mathcal{F}_k} & A_{\mathcal{F}_k\mathcal{A}_k} & 0\\
 A_{\mathcal{A}_k\mathcal{F}_k} & A_{\mathcal{A}_k\mathcal{A}_k} & B_{\mathcal{A}_k}\\
 0 & B_{\mathcal{A}_k} & 0
\end{pmatrix}
\begin{pmatrix}
 U^k_{\mathcal{F}_k}\\ U^k_{\mathcal{A}_k}\\ \Lambda^k_{\mathcal{A}_k}
\end{pmatrix}
=
\begin{pmatrix}
 F_{\mathcal{F}_k}\\ F_{\mathcal{A}_k}\\ G_{\mathcal{A}_k}
\end{pmatrix}.
@f}
This shows that one in fact only needs to solve for the Lagrange multipliers
located on the active set. By considering the second row one would then recover
the full Lagrange multiplier vector through
@f{equation*}
 \Lambda^k_S = B^{-1}\left(f_{\mathcal{S}} - A_{\mathcal{S}}U^k_{\mathcal{S}}\right).
@f}
Because of the third row and the fact that $B_{\mathcal{A}_k}$ is a diagonal matrix we are able
to calculate $U^k_{\mathcal{A}_k}=B^{-1}_{\mathcal{A}_k}G_{\mathcal{A}_k}$ directly. We can therefore also write the
linear system as follows:
@f{equation*}
\begin{pmatrix}
 A_{\mathcal{F}_k\mathcal{F}_k} & 0\\
 0 & Id_{\mathcal{A}_k} \\
\end{pmatrix}
\begin{pmatrix}
 U^k_{\mathcal{F}_k}\\ U^k_{\mathcal{A}_k}
\end{pmatrix}
=
\begin{pmatrix}
 F_{\mathcal{F}_k} - A_{\mathcal{F}_k\mathcal{A}_k}B^{-1}_{\mathcal{A}_k}G_{\mathcal{A}_k}
 \\
 B_{\mathcal{A}_k}^{-1}G_{\mathcal{A}_k}
\end{pmatrix}.
@f}
Fortunately, this form is easy to arrive at: we simply build the usual Laplace
linear system
@f{equation*}
\begin{pmatrix}
 A_{\mathcal{F}_k\mathcal{F}_k} & A_{\mathcal{F}_k\mathcal{A}_k} \\
 A_{\mathcal{A}_k\mathcal{F}_k} & A_{\mathcal{A}_k\mathcal{A}_k}
\end{pmatrix}
\begin{pmatrix}
 U^k_{\mathcal{F}_k}\\ U^k_{\mathcal{A}_k}
\end{pmatrix}
=
\begin{pmatrix}
 F_{\mathcal{F}_k}\\ F_{\mathcal{A}_k}
\end{pmatrix},
@f}
and then let the AffineConstraints class eliminate all constrained degrees of
freedom, namely $U^k_{\mathcal{A}_k}=B^{-1}_{\mathcal{A}_k}G_{\mathcal{A}_k}$,
in the same way as if the dofs in $\mathcal{A}_k$ were Dirichlet data. The
result linear system (the second to last one above) is symmetric and positive
definite and we solve it with a CG-method
and the AMG preconditioner from Trilinos.


<h3>Implementation</h3>

This tutorial is quite similar to step-4. The general structure of the program
follows step-4 with minor differences:
- We need two new methods, <code>assemble_mass_matrix_diagonal</code> and
  <code>update_solution_and_constraints</code>.
- We need new member variables that denote the constraints we have here.
- We change the preconditioner for the solver.


You may want to read up on step-4 if you want to understand the
current program.
